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人気のある微分積分 >

integral from 1 to 4 of 1/(x+1+sqrt(x))

解

\int_{1}^{4} \frac{1}{x + 1 + x} d x

解

\frac{3 3 ln ( \frac{7}{3} ) - 2 ( 3 arctan ( \frac{5 3}{3} ) - π )}{3 3}

+1

十進法表記

0.62775 \dots

解答ステップ

\int_{1}^{4} \frac{1}{x + 1 + x} d x

u置換積分法を適用する

= \int_{1}^{2} \frac{2 u}{u ^{2} + 1 + u} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{2} \frac{u}{u ^{2} + 1 + u} d u

平方完成する

u^{2} + 1 + u : (u + \frac{1}{2})^{2} + \frac{3}{4}

= 2 \cdot \int_{1}^{2} \frac{u}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

u置換積分法を適用する

= 2 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 ( 2 v - 1 )}{4 v ^{2} + 3} d v

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot 2 \cdot \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v - 1}{4 v ^{2} + 3} d v

拡張

\frac{2 v - 1}{4 v ^{2} + 3} : \frac{2 v}{4 v ^{2} + 3} - \frac{1}{4 v ^{2} + 3}

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot 2 (\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v}{4 v ^{2} + 3} d v - \int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{4 v ^{2} + 3} d v)

\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{2 v}{4 v ^{2} + 3} d v = \frac{1}{4} ln (\frac{7}{3})

\int_{\frac{3}{2}}^{\frac{5}{2}} \frac{1}{4 v ^{2} + 3} d v = \frac{3 arctan ( \frac{5}{3} ) - π}{6 3}

= 2 \cdot 2 \frac{1}{4} ln (\frac{7}{3}) - \frac{3 arctan ( \frac{5}{3} ) - π}{6 3}

簡素化

2 \cdot 2 \frac{1}{4} ln (\frac{7}{3}) - \frac{3 arctan ( \frac{5}{3} ) - π}{6 3} : \frac{3 3 ln ( \frac{7}{3} ) - 2 ( 3 arctan ( \frac{5 3}{3} ) - π )}{3 3}

= \frac{3 3 ln ( \frac{7}{3} ) - 2 ( 3 arctan ( \frac{5 3}{3} ) - π )}{3 3}

グラフ

人気の例

integral from-5 to-1 of 1/16

\int_{- 5}^{- 1} \frac{1}{16} d x

integral from 0 to pi/2 of e^{-st}sin(t)

\int_{0}^{\frac{π}{2}} e^{- s t} sin (t) d t

integral from 0 to 5 of (x^2+4)

\int_{0}^{5} (x^{2} + 4) d x

integral from 1 to 2 of (x-4)

\int_{1}^{2} (x - 4) d x

integral from-pi to pi of sin(2x)cos(x)

\int_{- π}^{π} sin (2 x) cos (x) d x